How to find odds of flooping....

[adam5206]

How do u calculate your odds preflop of flopping something. Say you have a pair how do u find the odds of flopping a set?

[man]

I'd recommend a probability course. but to put it briefly, you do it by counting the number of flops that will give you whatever hand you're looking for.

to find three of a kind: if you're holding two random cards, say 98, you want to find out how many flops will contain two 9's or two 8's, and then you divide by the total number of flops possible. so, we'll figure out how many flops will give you three 9's, and then multiply that number by 2 to get the total number for sets of 8's and sets of 9's.

if you have the 8/images/graemlins/spade.gif there are 3 8's left in the deck. so, your three of a kind can look like any of these (X is any random card that is not an 8):

8/images/graemlins/diamond.gif 8/images/graemlins/heart.gif X
8/images/graemlins/heart.gif 8 /images/graemlins/club.gif X
8/images/graemlins/club.gif 8/images/graemlins/diamond.gif X

so there are three combinations of 8's that we can have in our flop. we multiply this by the 44 other cards left in the deck (those that are not 8's and 9's, so that we don't double-count four of a kind and full houses), and get 132 (do you see why we multiply by 44?). multiply this by 2 to account for sets of nines, and we have 264 flops that make us happy. divide this by the 19600 possible flops, and we have the probability of hitting three of a kind on the flop as .013469 (1.3469%).


I didn't go into detail about the functions used in counting, most notably the combinatoric method (or whatever it's called). poke around at www.mathworld.com (http://www.mathworld.com) if you're really curious. I'll give a brief explanation here.

if you have a collection of items (like cards) and you want to pick out a certain number of them (like five, for a poker hand), you can count them by using this formula. let n stand for the number of available items (52 cards in the deck, for example), and let k stand for the number of items in each set (5 cards in each poker hand).

n!/(k!(n - k)!)

around here we use C(n,k) as shorthand for this formula.

So, to get the total number of possible poker hands, we calculate:

52!/(5!(52-5)!) = 52!/(5!(47!)) = (here's where my calculator craps out. darn.)

anyway, this won't make much sense unless you spend some time with a textbook. but if you want to learn it just takes practice. I feel like what I've written is going to sound meaningless if you don't already know what I'm talking about, but meh. hope this at least gets things started.

[man]

oh and haha, you spelled "flopping" wrong.

[LetYouDown]

For the question you actually asked...when you hold a pair, your odds are as follows:

C(50,3) possible flops or 19600.

C(2,1) ways of choosing your set card

C(12,2) ways of choosing two other cards that aren't paired (so no full house, no four of a kind).

16 = number of ways the suits can fall.

So. C(2,1) * C(12,2) * 16 = 2112, out of 19600. Which translates to a roughly 10.78% that you will flop a set, and only a set, with a pocket pair.